What is the integral from 0 to 1 of xsqrt(x-1) ?

The integral from 0 to 1 of xsqrt(x-1) is (4i)/(15)

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integral from 0 to 1 of xsqrt(x-1)

\int_{0}^{1} x x - 1 d x

\frac{4 i}{15}

Solution steps

\int_{0}^{1} x x - 1 d x

Apply u-substitution

: \int_{- 1}^{0} (u + 1) u d u

= \int_{- 1}^{0} (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

= \int_{- 1}^{0} u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{0} u^{\frac{3}{2}} d u + \int_{- 1}^{0} u d u

\int_{- 1}^{0} u^{\frac{3}{2}} d u = - \frac{2 i}{5}

\int_{- 1}^{0} u d u = \frac{2 i}{3}

= - \frac{2 i}{5} + \frac{2 i}{3}

Simplify

- \frac{2 i}{5} + \frac{2 i}{3} : \frac{4 i}{15}

= \frac{4 i}{15}

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What is the integral from 0 to 1 of xsqrt(x-1) ?
The integral from 0 to 1 of xsqrt(x-1) is (4i)/(15)